How far would a spacecraft moving in a circular orbit 500 km above Pluto's surface travel during that time?
1 answer:
Answer:
The speed of the spacecraft should be 719.35m/s
Explanation:
if the spacecraft is orbiting the planet with a circular orbit, the gravitational force must act as a centripetal force. This means:
In this case, the pluto's mass M is 1.3099·10^22 kg. The radius of the planet R is 1188.3Km and G is the gravitational constant. Therefore:
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Decreases
Explanation:
The force of attraction between two objects will decrease as the distance between them increases.
This is in compliance with newtons law of universal gravitation.
The force of attraction between the two bodies is a gravitational force.
According to newton's law of universal gravitation "the force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".
We see that from the last statement, the force of attraction between is inversely proportional to the square of the distances between them.
F ∞
r is the distance between them.
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Universal gravitation brainly.com/question/1724648
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Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s
Explanation:
hope it helps !!!!!!!!!!!!!
